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antoniya [11.8K]
3 years ago
7

Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra

mp and the block on the ramp is 0.36. You may assume the tension in the string connecting the two blocks is the same for all parts of the string. The two arrows define a coordinate system for the block on the ramp. The x axis is parallel to the ramp, the y axis is perpendicular to the ramp. The angle the ramp makes with the horizontal ground is 35 degrees. The block on the ramp has a weight of 13 Newtons.
Required:
What is the heaviest weight that can be hung below the pulley without the block on the ramp moving?
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

                                                  F = ma

   T     -   Wt ||      -         Ff            = m(0)

  mg   -  Μgsinθ -      μΜgcosθ    = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0        

                                                 m = 13(sin35 + 0.36cos35) / 9.8

                                                 m = 1.15205... ≈ 1.2 kg

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8 0
3 years ago
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​
Paraphin [41]

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.

6 0
3 years ago
When a car of mass 1200 kg, going with speed 30 m/s, rounds an unbanked curve of radius 150 m, what is the minimum coefficient o
Nana76 [90]

Answer:

\mu_s=0.61

Explanation:

In order for the car does not slip, the frictional force must be equal to the centripetal force due to the circular motion. According to the free body diagram:

\sum F_y:N=mg\\\sum F_x:F_f=F_c

The frictional force is given by:

F_f=\mu_s N=\mu_s mg

The centripetal force is defined as:

F_c=ma_c=m\frac{v^2}{r}

Here v is the linear speed and r is the radius of the circular motion. Replacing this equations:

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8 0
4 years ago
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